Simplifying Negative Exponents: (a^4/b^2)^-8

$(\frac{a^4}{b^2})^{-8}=\text{?}$

To solve this problem, we'll simplify the expression using exponent rules:

• Step 1: Apply the negative exponent rule.
  The expression $\left(\frac{a^4}{b^2}\right)^{-8}$ with a negative exponent can be rewritten using the reciprocal:
  $\left(\frac{a^4}{b^2}\right)^{-8} = \frac{1}{\left(\frac{a^4}{b^2}\right)^{8}}$
• Step 2: Apply the power of a quotient rule.
  Using $\left(\frac{x}{y}\right)^n = \frac{x^n}{y^n}$, we have:
  $\frac{1}{\left(\frac{a^4}{b^2}\right)^{8}} = \frac{1}{\frac{(a^4)^8}{(b^2)^8}} = \frac{1}{\frac{a^{32}}{b^{16}}}$
• Step 3: Simplify by taking the reciprocal of the fraction.
  Thus, $\frac{1}{\frac{a^{32}}{b^{16}}} = \frac{b^{16}}{a^{32}}$, which can be rewritten using exponent rules as $b^{16}a^{-32}$.

Therefore, the simplified expression is $b^{16}a^{-32}$.

Given the multiple-choice options, the correct choice is the one that matches our final result.

Therefore, the solution to the problem is $b^{16}a^{-32}$.